First slide

Heat engine

Question

The efficiency of Carnot engine is 50% and temperature of sink is 500 K. If the temperature of source is kept constant and its efficiency is to be raised to 60% then the required temperature of sink will be

Easy

A

600 K
B

500 K
C

400 K
D

100 K

Solution

$\begin{aligned} 1 - \frac{T_{2}}{T_{1}} = \frac{50}{100} \Rightarrow T_{1} = 2 T_{2} = 2 \times 500 k = 1000 k \\ ∴ 1 - \frac{T_{2}^{1}}{1000} = \frac{60}{100} \Rightarrow T_{2}^{1} = 0 .4 \times 1000 k = 400 K \end{aligned}$

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